Answer:
3.0%
Explanation:
A solution's mass percent concentration, (% m/m), is calculated by observing the mass of solute in grams of obtained in each 100 g of solution. In simple terms, the solution's mass percent concentration is given by the number of parts of solute obtained in every 100 parts of solution, made up of solute and solvent.
Hence the percent m/m concentration of a solution is obtained as;
Mass of solute/ mass of solution × 100/1
From the question, we have,
Mass of solute= 1.4g
Mass of solution= 46.6 g
Therefore, percent m/m concentration of the saline solution=
1.4/46.6 ×100/1 = 3.0%
The percent solute concentration of the solution is 3.0%
Answer:
The correct answer is B.
The 
is samller than 
of the reaction . So,the reaction will shift towards the left i.e. towards the reactant side.
Explanation:
Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as
K is the constant of a certain reaction when it is in equilibrium, while Q is the quotient of activities of products and reactants at any stage other than equilibrium of a reaction.
For the given chemical reaction:
The expression for is written as:
![Q=\frac{[PCl_3][Cl_2]}{[[PCl_5]^1}](https://tex.z-dn.net/?f=Q%3D%5Cfrac%7B%5BPCl_3%5D%5BCl_2%5D%7D%7B%5B%5BPCl_5%5D%5E1%7D)
Given : = 0.0454
Thus as 
, the reaction will shift towards the left i.e. towards the reactant side.
Answer:
7.78×10¯³ mole
Explanation:
From the question given above, the following data were obtained:
Volume (V) = 75 mL
Pressure (P) = 255 kPa
Temperature (T) = 22.5 °C
Number of mole (n) =?
Next, we shall convert 75 mL to L. This can be obtained as follow:
1000 mL = 1 L
Therefore,
75 mL = 75 mL × 1 L / 1000 mL
75 mL = 0.075 L
Next, we shall convert 22.5 °C to Kelvin temperature. This can be obtained as follow:
T(K) = T(°C) + 273
Temperature (T) = 22.5 °C
Temperature (T) = 22.5 °C + 273
Temperature (T) = 295.5 K
Finally, we shall determine the amount of oxygen molecules present in steel calorimeter. This can be obtained as follow:
Volume (V) = 0.075 L
Pressure (P) = 255 kPa
Temperature (T) = 295.5 K
Gas constant (R) = 8.314 KPa.L/Kmol
Number of mole (n) =?
PV = nRT
255 × 0.075 = n × 8.314 × 295.5
19.125 = n × 2456.787
Divide both side by 2456.787
n = 19.125 / 2456.787
n = 7.78×10¯³ mole
Thus, amount of oxygen molecules present in steel calorimeter is 7.78×10¯³ mole
Answer:
262.5 mL
Explanation:
21% of 1.25 L volume is methanol, so the amount of methanol is,
0.21 × 1250 mL = 262.5 mL
<h2>Answer:</h2>
Option B is correct.
b. extending outward is more economical than extending upward
<h2>Explanation:</h2>
people tend to move from rural areas to urban for the greater facilities that they can enjoy at cities. This urban sprawl is also very economic because people will get more chances of jobs, business and work hence improving the economy.
